"Polygons and gravitons" and Kodaira's theorem I'm trying to understand the paper by Hitchin called: ''Polygons and gravitons". I'm stuck at page 470. At this point, he does some computations and obtains the conformal structure of the real manifold.
I know that because of Kodaria's theorem, there exist a 4-dimensional complex manifold that parametrizes the sections of $H^k\oplus H^k\oplus H^2$. Furthermore, because of this theorem, there is an isomorphism between the sections of the normal bundle and the tangent vectors of the 4-dimensional manifold. Using the notation of the article, a point $(a,b,c,A)$ of the manifold corresponds to a section of $H^k\oplus H^k\oplus H^2$, i.e. to certain $x(u),y(u),z(u)$. However, I don't understand how to see the relation between a tangent vector $(a',b',c',A')$ on the point $(a,b,c,A)$ and a section $(x',y',z')\in ker(f_x,f_y,f_z)$.
Actually, I also don't know how to interpret this section $(x',y',z')$. Is it a section of $H \oplus H$ ?
I can follow the computations done at this point but I don't see this equality:
$$Re\left(\frac{2A'}{A} \right)=\sum \frac{(b-b_i)'+\Delta_i'}{(b-b_i)+\Delta_i}=\gamma b'+Re(\delta a')$$
Any help will be very welcome. Thank you!
 A: The computations are these:
Let us see this equality. 
$$\ln\left(A\bar{A}\right)=\ln A+\ln \bar{A}=2\text{Re}\ln A =\ln\left(\prod \left((b-b_i)-\Delta_i \right)\right)=\sum \ln\left((b-b_i)-\Delta_i \right)\Rightarrow$$
$$\Rightarrow (2\text{Re}\ln A)'=2Re\left(\dfrac{A'}{A}\right)=Re\left(\dfrac{2A'}{A}\right)=\sum \dfrac{(b-b_i)'+\Delta_i'}{(b-b_i)+\Delta_i}$$
On the other hand, 
$$\dfrac{|a-a_i|'}{|a-a_i|}=(\ln|a-a_i|)'=\dfrac{1}{2}\ln\left((a-a_i)(\bar{a}-\bar{a}_i)\right)'=\dfrac{1}{2}\left( 2\text{Re}\ln (a-a_i) \right)'=\text{Re}\left( \dfrac{a'}{a-a_i}\right)$$
Hence, using that $\alpha_i=\frac{-(b-b_i)+\Delta_i}{a-a_i}$
$$\text{Re}(\delta a')=\sum \text{Re}\left(\dfrac{a'}{a-a_i}\dfrac{\Delta_i-(b-b_i)}{\Delta_i} \right)=\sum \text{Re}\left(\dfrac{a'}{a-a_i}\right)\dfrac{\Delta_i-(b-b_i)}{\Delta_i}\Rightarrow $$$$\Rightarrow \text{Re}(\delta a') = \sum\dfrac{|a-a_i|'}{|a-a_i|}\dfrac{\Delta_i-(b-b_i)}{\Delta_i}$$
and therefore we finally obtain:
$$\text{Re}(\delta a')+\gamma b'=\sum\left( \dfrac{|a-a_i|'}{|a-a_i|}\dfrac{\Delta_i-(b-b_i)}{\Delta_i}+\dfrac{b'}{\Delta_i}\right)=$$$$=\sum \left(\dfrac{\dfrac{|a-a_i|'}{|a-a_i|}\left(\Delta_i-\Delta_i^{-1}(b-b_i)^2\right)+b'+\Delta_i^{-1}(b-b_i)b'}{(b-b_i)+\Delta_i} \right)=$$$$=\sum \left(\dfrac{\dfrac{|a-a_i|'}{|a-a_i|}\dfrac{|a-a_i|^2}{\Delta_i}+b'+\dfrac{(b-b_i)b'}{\Delta_i}}{(b-b_i)+\Delta_i} \right)= $$$$=\sum \left(\dfrac{\dfrac{|a-a_i||a-a_i|'+(b-b_i)b'}{\Delta_i}+b'}{(b-b_i)+\Delta_i} \right)=\sum\frac{(b-b_i)'+\Delta_i'}{(b-b_i)+\Delta_i}$$
